Question

If $f\left(x\right)=0$ is a quadratic equation such that $f(-\pi )=f\left(\pi \right)=0$ and $f\left(\frac{\pi }{2}\right)=-\frac{3{\pi }^2}{4}$, then ${lim}_{x\to -\pi }\frac{f\left(x\right)}{\sin \left(\sin x\right)}$ is

• A. 0
• B. $\pi$
• C. $2\pi$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C

$2\pi$

Given :

$f(-\pi )=f\left(\pi \right)=0$ and $f\left(\frac{\pi }{2}\right)=-\frac{3{\pi }^2}{4}$

$\therefore f\left(x\right)=x^2-{\pi }^2$
$\Rightarrow {lim}_{x\to -\pi }\frac{x^2-{\pi }^2}{\sin \left(\sin x\right)}$

$={lim}_{h\to 0}\frac{(-\pi +h{)}^2-{\pi }^2}{\sin \left(\sin \right(-\pi +h\left)\right)}$

$={lim}_{h\to 0}\frac{-2h\pi +h^2}{-\sin \left(\sin h\right)}$
$={lim}_{h\to 0}\frac{h-2\pi }{\frac{-\sin \left(\sin h\right)}{\sinh }\times \frac{\sin h}{h}}=2\pi$